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%I #15 Sep 23 2022 17:02:20
%S 1,1,2,10,70,614,6694,86950,1306238,22301182,426568582,9034270022,
%T 209865005182,5305633245502,145015267113254,4261031552171302,
%U 133942497987918142,4485091167113782334,159382471398546619270,5990690461349053361350,237457043901226772247998
%N E.g.f. satisfies: A(x) = exp( Integral 1 + x*A(x)^3 dx ), where the constant of integration is zero.
%C Compare to the identities:
%C (1) F(x) = exp( Integral 1 + x*F(x) dx ) when F(x) = 1/(1-x),
%C (2) G(x) = exp( Integral x*G(x)^3 dx ) when G(x) = 1/(1-3*x^2/2)^(1/3).
%C In general, if e.g.f. satisfies: A(x) = exp( Integral(1 + x*A(x)^p) dx ), p>1, and the constant of integration is zero, then A(x) = (1/p + (p-1)/(exp(p*x)*p) - x)^(-1/p), and a(n) ~ n! * p^(n+1/p) / (Gamma(1/p) * n^(1-1/p)* (1+LambertW((p-1)*exp(-1)))^(n+2/p)). - _Vaclav Kotesovec_, Jul 16 2014
%F E.g.f.: 3^(1/3)*exp(x)/(exp(3*x) - 3*exp(3*x)*x + 2)^(1/3). - _Vaclav Kotesovec_, Jan 05 2014
%F a(n) ~ 3^(n+5/6) * n^(n-1/6) * Gamma(2/3) / (sqrt(2*Pi) * exp(n) * (1+LambertW(2*exp(-1)))^(n+2/3)). - _Vaclav Kotesovec_, Jan 05 2014
%e E.g.f.: A(x) = 1 + x + 2*x^2/2! + 10*x^3/3! + 70*x^4/4! + 614*x^5/5! + 6694*x^6/6! + ...
%e such that, by definition,
%e log(A(x)) = x + x^2/2! + 6*x^3/3! + 36*x^4/4! + 288*x^5/5! + 2970*x^6/6! + 36612*x^7/7! + ...
%e Related expansions:
%e x*A(x)^3 = x + 6*x^2/2! + 36*x^3/3! + 288*x^4/4! + 2970*x^5/5! + 36612*x^6/6! + ...
%e A(x)^3 = 1 + 3*x + 12*x^2/2! + 72*x^3/3! + 594*x^4/4! + 6102*x^5/5! + 75006*x^6/6! + ...
%t CoefficientList[Series[3^(1/3)*E^x/(E^(3*x) - 3*E^(3*x)*x + 2)^(1/3), {x, 0, 20}], x] * Range[0, 20]! (* _Vaclav Kotesovec_, Jan 05 2014 *)
%o (PARI) {a(n)=local(A=1+x);for(i=1,n,A=exp(intformal(1+x*A^3)+x*O(x^n)));n!*polcoeff(A,n)}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A212913, A245266, A245267.
%K nonn
%O 0,3
%A _Paul D. Hanna_, May 30 2012
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